Solve (a) $f(x) = 0$, (b) $f(x) > 0$, and (c) $f(x) < 0$ over the interval $[0, 2\pi)$.
$f(x) = 2\sin^2 x + \sin x$
The solution set for $f(x) = 0$ in the interval $[0, 2\pi)$ is $\boxed{}$
(Simplify your answer. Type an exact answer, using $\pi$ as needed. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.)
Find the solution set for $f(x) > 0$ in the specified interval. Choose the correct answer below.
A. The solution set of $f(x) > 0$ is $(\pi, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi)$.
B. The solution set of $f(x) > 0$ is $(\pi, \frac{7\pi}{6})$.
C. The solution set of $f(x) > 0$ is $(0, \pi) \cup (\frac{7\pi}{6}, \frac{11\pi}{6})$.
Find the solution set for $f(x) < 0$ in the specified interval. Choose the correct answer below.
A. The solution set of $f(x) < 0$ is $(0, \pi)$.
B. The solution set of $f(x) < 0$ is $(\frac{\pi}{6}, \frac{7\pi}{6})$.
C. The solution set of $f(x) < 0$ is $(\frac{\pi}{6}, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi)$.
